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Q.

A particle is executing SHM with time period π sec along a straight line. At t=0, it has started its motion from one of its extreme position. Then the minimum time after which its kinetic energy will be equal to its potential energy is

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Detailed Solution

Equation of SHM is  x=Acosωt At time t , potential energy,  U=12mω2x2=12mω2A2cos2ωtAnd kinetic energy, K=12mω2(A2−x2)=12mω2A2sin2ωtWhen K=U, 12mω2A2sin2ωt=12mω2A2cos2ωt ⇒tanωt=±1⇒ωt=π4,3π4,5π4,7π4,........etc∴tmin=π4ω=π4×2ππ=π8sec
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A particle is executing SHM with time period π sec along a straight line. At t=0, it has started its motion from one of its extreme position. Then the minimum time after which its kinetic energy will be equal to its potential energy is